HDU1074-Doing Homework

本文介绍了一个通过二进制状态压缩的动态规划算法解决课程作业最优排序的问题。目标是最小化因延期提交作业而被扣除的成绩分数,并确保在多种可行方案中选择字典序最小的作业完成顺序。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

Doing Homework

                                                                       Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
                                                                                                  Total Submission(s): 8147    Accepted Submission(s): 3744

Problem Description
Ignatius has just come back school from the 30th ACM/ICPC. Now he has a lot of homework to do. Every teacher gives him a deadline of handing in the homework. If Ignatius hands in the homework after the deadline, the teacher will reduce his score of the final test, 1 day for 1 point. And as you know, doing homework always takes a long time. So Ignatius wants you to help him to arrange the order of doing homework to minimize the reduced score.
 
Input
The input contains several test cases. The first line of the input is a single integer T which is the number of test cases. T test cases follow.
Each test case start with a positive integer N(1<=N<=15) which indicate the number of homework. Then N lines follow. Each line contains a string S(the subject's name, each string will at most has 100 characters) and two integers D(the deadline of the subject), C(how many days will it take Ignatius to finish this subject's homework). 

Note: All the subject names are given in the alphabet increasing order. So you may process the problem much easier.
 
Output
For each test case, you should output the smallest total reduced score, then give out the order of the subjects, one subject in a line. If there are more than one orders, you should output the alphabet smallest one.
 
Sample Input
2 3 Computer 3 3 English 20 1 Math 3 2 3 Computer 3 3 English 6 3 Math 6 3
 
Sample Output
2 Computer Math English 3 Computer English Math
Hint
In the second test case, both Computer->English->Math and Computer->Math->English leads to reduce 3 points, but the word "English" appears earlier than the word "Math", so we choose the first order. That is so-called alphabet order.
 
Author
Ignatius.L

       题意:有n门课,每门课有截止时间和完成所需的时间,课程按字典序输入,如果超过规定时间完成,每超过一天就会扣1分,问怎样安排做作业的顺序才能使得所扣的分最小,如果答案有多个按字典序输出

       思路:使用二进制来表示所有完成的状况

#include <iostream>
#include <string>
#include <cstring>
#include <stack>
#include <algorithm>
#include <stdio.h>

using namespace std;

const int INF=1<<30;

struct node
{
    string name;
    int dead,cost;
}a[50];

struct kode
{
    int time,score,pre,now;
}dp[1<<15];

int main()
{
    int t,n,en;
    scanf("%d",&t);
    while(t--)
    {
        memset(dp,0,sizeof(dp));
        scanf("%d",&n);
        for(int i=0; i<n; i++)
            cin>>a[i].name>>a[i].dead>>a[i].cost;
        en=1<<n;
        for(int i=1; i<en; i++)
        {
            dp[i].score=INF;
            for(int j=n-1; j>=0; j--)
            {
                int tem=1<<j;
                if(i&tem)
                {
                    int past=i-tem;
                    int st=dp[past].time+a[j].cost-a[j].dead;
                    if(st<0) st=0;
                    if(st+dp[past].score<dp[i].score)
                    {
                        dp[i].score=st+dp[past].score;
                        dp[i].now=j;
                        dp[i].pre=past;
                        dp[i].time=dp[past].time+a[j].cost;
                    }
                }
            }
        }
        stack<int> q;
        int temp=en-1;
        printf("%d\n",dp[temp].score);
        while(temp)
        {
            q.push(dp[temp].now);
            temp=dp[temp].pre;
        }
        while(!q.empty())
        {
            cout<<a[q.top()].name<<endl;
            q.pop();
        }
    }
    return 0;
}
评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值